给定大小为N的数组arr [] 。任务是检查是否存在大小至少为2的子数组，以使其总和为回文。如果存在这样的子数组，则打印YES 。否则，打印NO 。
例子：
Input: arr[] = {10, 6, 7, 9, 12}
Output: Yes
Explanation:
The subarray [6, 7, 9] with sum 22 is palindrome.
Input: arr[] = {15, 4, 8, 2}
Output: No
Explanation:
No such subarray exists.
方法：
要解决此问题，请执行以下步骤：
- 创建给定数组的前缀和数组。
- 使用嵌套的for循环遍历数组，以表示子数组的开始和结束。索引[x，y]中子数组的总和可以通过pref [y] – pref [x – 1]获得。
- 检查此总和是否是回文。如果是回文，则打印“是”，否则打印“否”。
下面是上述方法的实现：
C++
// C++ program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
#include
using namespace std;
// Function which checks whether
// a given number is palindrome or not
bool checkPalindrome(int n)
{
// Store the reverse of
// the number n
int rev = 0;
for (int x = n; x != 0; x /= 10) {
int d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
void findSubarray(int ar[], int n)
{
// Making a prefix sum array of ar[]
int pref[n];
pref[0] = ar[0];
for (int x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
bool found = false;
for (int x = 0; x < n; x++) {
for (int y = x + 1; y < n; y++) {
// sum stores the sum of subarray
// from index x to y of array
int sum = pref[y];
if (x > 0) {
sum -= pref[x - 1];
}
if (checkPalindrome(sum)) {
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
// Driver code
int main()
{
int ar[] = { 1, 11, 20, 35 };
int n = sizeof(ar) / sizeof(ar[0]);
findSubarray(ar, n);
return 0;
} Java
// Java program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
class GFG{
// Function which checks whether
// a given number is palindrome or not
static boolean checkPalindrome(int n)
{
// Store the reverse of
// the number n
int rev = 0;
for(int x = n; x != 0; x /= 10)
{
int d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
static void findSubarray(int []ar, int n)
{
// Making a prefix sum array of ar[]
int []pref = new int[n];
pref[0] = ar[0];
for(int x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
boolean found = false;
for(int x = 0; x < n; x++)
{
for(int y = x + 1; y < n; y++)
{
// sum stores the sum of subarray
// from index x to y of array
int sum = pref[y];
if (x > 0)
{
sum -= pref[x - 1];
}
if (checkPalindrome(sum))
{
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
System.out.println("Yes");
else
System.out.println("No");
}
// Driver code
public static void main(String args[])
{
int []ar = { 1, 11, 20, 35 };
int n = ar.length;
findSubarray(ar, n);
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to check if sum of
# any subarray of size atleast 2 is
# palindrome or not
# Function which checks whether a
# given number is palindrome or not
def checkPalindrome(n):
# Store the reverse
# of the number n
rev = 0
x = n
while(x != 0):
d = x % 10
rev = rev * 10 + d
x = x // 10
if (rev == n):
return True
else:
return False
# Function which checks if the
# requires subarray exists or not
def findSubarray(ar, n):
# Making a prefix sum array of ar[]
pref = [0 for i in range(n)]
pref[0] = ar[0]
for x in range(1, n):
pref[x] = pref[x - 1] + ar[x]
# Boolean variable that will store
# whether such subarray exists or not
found = False
for x in range(n):
for y in range(x + 1, n, 1):
# Sum stores the sum of subarray
# from index x to y of array
sum = pref[y]
if (x > 0):
sum -= pref[x - 1]
if (checkPalindrome(sum)):
# Required subarray is found
found = True
break
if (found):
break
if (found):
print("Yes")
else:
print("No")
# Driver code
if __name__ == '__main__':
ar = [ 1, 11, 20, 35 ]
n = len(ar)
findSubarray(ar, n)
# This code is contributed by Surendra_GangwarC#
// C# program to check if sum of any
// subarray of size atleast 2 is
// palindrome or not
using System;
class GFG{
// Function which checks whether
// a given number is palindrome or not
static bool checkPalindrome(int n)
{
// Store the reverse of
// the number n
int rev = 0;
for(int x = n; x != 0; x /= 10)
{
int d = x % 10;
rev = rev * 10 + d;
}
if (rev == n)
return true;
else
return false;
}
// Function which checks if the
// requires subarray exists or not
static void findSubarray(int []ar, int n)
{
// Making a prefix sum array of ar[]
int []pref = new int[n];
pref[0] = ar[0];
for(int x = 1; x < n; x++)
pref[x] = pref[x - 1] + ar[x];
// Boolean variable that will store
// whether such subarray exists or not
bool found = false;
for(int x = 0; x < n; x++)
{
for(int y = x + 1; y < n; y++)
{
// sum stores the sum of subarray
// from index x to y of array
int sum = pref[y];
if (x > 0)
{
sum -= pref[x - 1];
}
if (checkPalindrome(sum))
{
// Required subarray is found
found = true;
break;
}
}
if (found)
break;
}
if (found)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver code
public static void Main()
{
int []ar = { 1, 11, 20, 35 };
int n = ar.Length;
findSubarray(ar, n);
}
}
// This code is contributed by Code_Mech输出：
Yes
时间复杂度： O(N 2 )
辅助空间： O(1)